How to find in which number base the operation was done by looking at the corresponding operation in decimal...
$begingroup$
$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this
Method 1 : Through equations
assume base be a
$$23_a + 25_a = 51_a $$
$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7
Method 2 : Just logical reasoning
as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .
No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .
To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?
That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .
I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .
Thanks
elementary-number-theory number-systems
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
$endgroup$
add a comment |
$begingroup$
$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this
Method 1 : Through equations
assume base be a
$$23_a + 25_a = 51_a $$
$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7
Method 2 : Just logical reasoning
as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .
No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .
To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?
That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .
I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .
Thanks
elementary-number-theory number-systems
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
$endgroup$
1
$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29
$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04
2
$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11
add a comment |
$begingroup$
$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this
Method 1 : Through equations
assume base be a
$$23_a + 25_a = 51_a $$
$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7
Method 2 : Just logical reasoning
as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .
No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .
To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?
That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .
I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .
Thanks
elementary-number-theory number-systems
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
$endgroup$
$$23 + 25 = 51 $$ What base is used in the above addition operation ?
I have 2 methods to do this
Method 1 : Through equations
assume base be a
$$23_a + 25_a = 51_a $$
$$2a + 3 + 2a + 5 = 5a +1 $$
solving this we get a as 7
Method 2 : Just logical reasoning
as the digit in $ a^1 $'s place is $5$ instead of $4$ (because $2 + 2 =4$). This means that a carry came from addition of digits in $ a^0 $'s place . This means $ 5 + 3 = 11 $ here . so it means $ 5 + 1 = 6$ (Just finding addition which does not result in carry) . so base is $7$ because after $6$ adding $1$ leads to $10$ , then another $1$ added leads to $11$ .
No problem with the above methods . But these methods are usable and easy to visualise when numbers are small and when simple operations like addition are involved . The methods become tedious to follow when number's are large and when product or division operations are given .
So is there a better way to approach these kind of problems where an operation is given and asked to find the base in which the operation was performed .
To be more specific and precise,Is there way to tell the base just by comparing the operation's result (in the unknown base) and value it will get in decimal system(base 10) and tell the base in which the operation was performed ?
That is as $$ 23_a + 25_a = 51_a $$
$$ 23_{10} + 25_{10} = 48_{10} $$
Comparing $48_{10}$ (result in decimal system) and $51_a$ (result in base a system) will I be able to conclude that base a is nothing but $7$ .
I am quiet jealous in wanting to find it by seeing its result in decimal system because in general we are more used to decimal system than any other base .
Thanks
elementary-number-theory number-systems
elementary-number-theory number-systems
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
asked Jul 26 '13 at 10:17
Harish KayarohanamHarish Kayarohanam
1,6701924
1,6701924
1
$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29
$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04
2
$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11
add a comment |
1
$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29
$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04
2
$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11
1
1
$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29
$begingroup$
If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
$endgroup$
– egreg
Jul 26 '13 at 11:29
$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04
$begingroup$
I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:04
2
2
$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11
$begingroup$
We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
$endgroup$
– egreg
Jul 26 '13 at 20:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
$endgroup$
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
add a comment |
$begingroup$
I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.
For your problem,
begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
$endgroup$
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
add a comment |
$begingroup$
In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
$endgroup$
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
add a comment |
$begingroup$
In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
$endgroup$
In general, you will find that you get expressions which convert to polynomial equations in the base. If you know the base is an integer, you will then be able to use the rational root theorem to test the limited number of possible bases. Some of these possibilities are likely to be eliminated by other considerations.
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
answered Jul 26 '13 at 11:22
Mark BennetMark Bennet
81.8k984183
81.8k984183
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
add a comment |
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
Hi @Mark Bennet , the equation way of solving has been mentioned in the question as a known method . And I have added that this was of solving becomes tedious for big numbers because we have equations in 3 rd degree or greater . That's why I asked if there is a logical way of doing it just by seeing the result in decimal system.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:07
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
@HarishKayarohanam How do you expect the answer to be anything else? Write the thing out and see what it implies. I have suggested a way of reducing the computation.
$endgroup$
– Mark Bennet
Jul 26 '13 at 20:09
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
$begingroup$
I am not suggesting the answer to be anything else , but the method to be something else .But your answer is still based on equation way of solving it.
$endgroup$
– Harish Kayarohanam
Jul 26 '13 at 20:12
add a comment |
$begingroup$
I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.
For your problem,
begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.
For your problem,
begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
$endgroup$
add a comment |
$begingroup$
I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.
For your problem,
begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
$endgroup$
I like to use the notation $[a,b]_B = aB + b, [a,b,c]_B=aB^2+bB + c$, and so on.
For your problem,
begin{align}
[2,3]_a + [2,5]_a &= [5,1]_a \
[4,8]_a &= [5,1]_a \
[4,8]_a &= [4, a+1] \
8 &= a+ 1 \
a &= 7
end{align}
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
answered Jan 1 at 8:40
steven gregorysteven gregory
18.3k32358
18.3k32358
add a comment |
add a comment |
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1
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If $b$ is the base ($b>5$, by implicit assumption), then $3+5=b+1$.
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– egreg
Jul 26 '13 at 11:29
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I didn't get your point . Also what is this assumption b> 5 .Could you please elaborate .
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– Harish Kayarohanam
Jul 26 '13 at 20:04
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We assume that the digits have the usual meaning, otherwise there would be no clue. Since there is the digit $5$, the base must be at least $6$; then the operation on the rightmost digit can carry at most $1$, and it does, because the rightmost digit in the sum is $1$. Hence $5+3=b+1$, so $b=7$. Then you can verify that the operation has been correctly performed, because $1+2+2=5$.
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– egreg
Jul 26 '13 at 20:11